The way I start is to see if I can spot an easy factor like x=1 or 2, because the polynomial can be simplified if you can. The way you do this is find out what value of x makes the expression zero. x=1 gives you a result -384. So that's way out. Try x=2. It gives zero! Great! Can we find any more by substitution? x=3 also gives zero. That's enough. It means that we have two factors (x-2) and (x-3). We can divide the polynomial by these in turn or in combination. If we go for combination, then the combination produces the quadratic x^2-5x+6. We now use algebraic long division to find out more factors. To divide the quadratic into the polynomial we look to see how many times x^2 goes into x^5. The answer is x^3, so we multiply the quadratic by this value and get x^5-5x^4+6x^3. (x^3 is the first term in the quotient.) Subtract this from the polynomial and we're left with 7x^4-60x^3-8x^2+725x-1050. This time 7x^2 is the required multiplier (the quotient is now x^3+7x^2.) Subtract 7x^4-35x^3+42x^2 from the remainder to get -25x^3-50x^2+725x-1050. Subtract -25x^3+125x^2-150x from the remainder to get -175x^2+875x-1050. (Qotient is now x^3+7x^2-25x.) Finally, we can subtract -175 times the quadratic leaving no remainder. The final quotient is x^3+7x^2-25x-175. We still need to factorise this. I'd go for x=5 because of the appearance of the numbers 25 and 175. In fact, x=5 is a factor when we substitute this value and get the cubic to zero. So we can divide the cubic by (x-5) and get x^2+12x+35. This factorises into (x+7)(x+5). Nice! So the total factorisation is (x-2)(x-3)(x-5)(x+7)(x+5).